Question

A uniform metal disc of radius $R$ is taken and out of it a disc of diameter $\frac{R}{2}$ is cut off from the end. The centre of mass of the remaining part will be:

• A. $\frac{R}{10}$ from the centre
• B. $\frac{R}{15}$ from the centre
• C. $\frac{R}{5}$ from the centre
• D. $\frac{R}{20}$ from the centre

Answer 1

The correct option is D $\frac{R}{20}$ from the centre

Density is uniform, so we can use geometrical method.

Area of disc, $A_1=\pi R^2$
Area of cutoff disc, $A_2=\frac{\pi }{4}{\left(\frac{R}{2}\right)}^2=\frac{\pi R^2}{16}$

$x_{com}=\frac{A_1\times 0-A_2{x}_2}{A_1-A_2}=\frac{-\frac{\pi R^2}{16}\times \frac{3R}{4}}{\pi R^2-\frac{\pi R^2}{16}}=\frac{-3R}{15\times 4}=\frac{-R}{20}$

Hence, center of mass is at $\frac{R}{20}$ from the center.